2.2.17Additional lessons
Hybridisation (HL) — IB Chemistry HL Revision Notes & Topical Questions
5 min read 6 parts Lesson 41 of 172
Part 1 of 6The Concept
Hybridisation is a model that explains molecular geometry using mixed atomic orbitals.
Hybridisation
- Mixing of atomic orbitals on one atom to form new orbitals of equal energy.
Number of hybrids
- The number of hybrid orbitals formed equals the number of atomic orbitals mixed.
Purpose
- Hybrid orbitals explain observed shapes, such as the four equal bonds in methane.
Link to domains
- The number of electron domains on the atom decides which hybridisation applies.
Part 2 of 6Types of Hybridisation
Types of Hybridisation
sp³
- One s and three p orbitals give four hybrids pointing to a tetrahedron, with angles of 109.5°.
sp²
- One s and two p orbitals give three hybrids in a plane at 120°, leaving one unhybridised p orbital.
sp
- One s and one p orbital give two hybrids at 180°, leaving two p orbitals.
Unhybridised p orbitals
- These form the pi bonds, so sp² atoms have one pi bond and sp atoms have two.
| Hybridisation | Domains | Shape | Example |
|---|---|---|---|
| sp³ | 4 | Tetrahedral | C in CH₄ |
| sp² | 3 | Trigonal planar | C in C₂H₄ |
| sp | 2 | Linear | C in C₂H₂ |
Part 3 of 6Assigning Hybridisation
1
Draw the Lewis structure and count electron domains on the atom of interest.
2
Four domains means sp³, three domains means sp² and two domains means sp.
3
Lone pairs count as domains, so nitrogen in NH₃ and oxygen in H₂O are sp³.
4
Each sigma bond and each lone pair uses a hybrid orbital; pi bonds use leftover p orbitals.
5
Check with the shape: tetrahedral arrangement indicates sp³.
Part 4 of 6Examples
Ethene
- Both carbons are sp² with a sigma framework and one pi bond.
Carbon dioxide
- The carbon is sp hybridised with two pi bonds.
Ethanoic acid
- The CH₃ carbon is sp³ and the carboxyl carbon is sp².
Propyne
- In CH₃–C≡CH the methyl carbon is sp³ and the two triple-bonded carbons are sp.
Part 5 of 6Worked Example
Question State the hybridisation of each carbon in propyne, CH₃–C≡C–H.
- CH₃ carbonfour sigma bonds, 4 domains, so sp³.
- Middle carbonone sigma bond to CH₃, one sigma bond and two pi bonds to the next carbon, so 2 domains, sp.
- Terminal carbonone sigma bond to H and one sigma plus two pi to carbon, so 2 domains, sp.
Answer sp³, sp, sp from left to right.
Part 6 of 6Exam Checklist & Tips
2.2.17 – Key Points for the Exam
- Hybridisation Mixing of atomic orbitals to form new equivalent orbitals.
- Hybrid orbital An orbital formed by combining s and p orbitals.
- sp² carbon A carbon with three sigma bonds and one pi bond.
- Unhybridised orbital A p orbital not used in mixing, available for pi bonding.
Common Mistakes
- Counting only bonds and ignoring lone pairs when assigning hybridisation.
- Saying that hybrid orbitals form pi bonds.
- Giving sp³ to carbon in a C=C bond.
- Forgetting that the number of hybrid orbitals equals the number mixed.
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